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devlian [24]
1 year ago
7

what is the difference between the sum of the first 20032003 even counting numbers and the sum of the first 20032003 odd countin

g numbers?
Mathematics
1 answer:
earnstyle [38]1 year ago
8 0

Answer: The difference between sum of first 20032003 even numbers and sum of first 20032003 odd numbers is “-20032003”.

Step-by-step explanation:

The sum of first 20032003 even numbers:

Summation E=E_{1}+E_{2}+E_{3}+ . . . . . +E_{20032003}

The sum of first 20032003 odd numbers:

Summation O=O_{1}+O_{2}+O_{3}+ . . . . . +O_{20032003}

Difference between the sum of first 20032003 even numbers and the sum of first 20032003 odd numbers:

Difference=Summation E - Summation O=(E_{1}+E_{2}+E_{3}+ . . . . . . . +E_{20032003})-(O_{1}+O_{2}+O_{3}+ . . . . . . . +O_{20032003})

Difference=(E_{1}-O_{1})+(E_{2}-O_{2})+(E_{3}-O_{3})+ . . . . . . . +(E_{20032003}-O_{20032003})

Even numbers start as 0, 2, 4, 6…. While odd numbers start as 1, 3, 5, 7…..

(assuming that we are taking non-negative even/odd numbers into account only)

As we know, every nth even number is one behind their respective odd number (i.e. 0 lags 1, 2 lags 3, 4 lags 5 and so on)

Thus, our equation of difference becomes,

Difference=(-1)_{1}+(-1)_{2}+(-1)_{3}+ . . . . . . . +(-1)_{20032003} =(-1)*20032003=-20032003

Learn more about even & odd numbers here brainly.com/question/24657250

#SPJ4

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a) p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

b) p_B represent the real population proportion for male

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c) (0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

Part b

p_B represent the real population proportion for male

\hat p_B =\frac{102}{170}=0.6 represent the estimated proportion for male

n_B=170 is the sample size elected for male

z represent the critical value for the margin of error  

Part c

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

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